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Second order equations questions

Guess e to the mx and a differential equation collapses to a quadratic. Its two roots, real, repeated or complex, dictate everything the solution can do; a right-hand side then adds one particular piece on top.

7 original questions · 29 marks · the second order equations notes · Differential equations

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• 2 marks · write · warm-up

  Write down the auxiliary equation for ay'' + by' + cy = 0, and state where it comes from.

  Worked answer

  am² + bm + c = 0. Substituting the trial solution y = e^mx puts a factor of e^mx in every term; cancelling it leaves the quadratic. B1 for the auxiliary equation, B1 for where it comes from.

• 3 marks · find · warm-up

  Find the general solution of y'' − 7y' + 12y = 0.

  Worked answer

  Auxiliary: m² − 7m + 12 = (m − 3)(m − 4) = 0, roots 3 and 4. General solution y = Ae^3x + Be^4x: two distinct real roots, two independent exponentials. M1 for the auxiliary equation, A1 for the roots, A1 for the general solution.

• 3 marks · find

  Find the general solution of y'' + 9y = 0, and state the period of every solution.

  Worked answer

  m² + 9 = 0 gives m = ±3i, so y = A cos 3x + B sin 3x. Pure oscillation at angular frequency 3, so every solution repeats with period 2π/3. M1 for m = ±3i, A1 for the general solution, B1 for the period.

• 3 marks · find

  Find the general solution of y'' − 6y' + 9y = 0.

  Worked answer

  m² − 6m + 9 = (m − 3)² = 0, a repeated root. The solution needs its extra factor of x, so y = (A + Bx)e^3x. M1 for the repeated root, A1 for the extra factor of x, A1 for the general solution. Without the Bx the two constants would collapse into one and two initial conditions could not be met.

• 6 marks

  Solve y'' − 2y' + 5y = 0 with y(0) = 2 and y'(0) = 0.

  Worked answer

  m² − 2m + 5 = 0 gives m = 1 ± 2i, so y = e^x(A cos 2x + B sin 2x). y(0) = 2 gives A = 2. Differentiating and setting x = 0 gives y'(0) = A + 2B = 0, so B = −1. The solution is y = e^x(2 cos 2x − sin 2x), an oscillation at frequency 2 growing inside e^x. M1 for the auxiliary equation, A1 for m = 1 ± 2i, A1 for the general solution, B1 for A = 2, M1 for differentiating and using y'(0) = 0, A1 for B = −1.

• 5 marks

  Solve y'' − y = 0 with y(0) = 1 and y'(0) = 0, and identify the solution as a standard function.

  Worked answer

  m² − 1 = 0 gives m = ±1, so y = Ae^x + Be^−x. The conditions give A + B = 1 and A − B = 0, so A = B = 1/2 and y = (e^x + e^−x)/2 = cosh x. M1 for the auxiliary equation, A1 for the general solution, M1 for both conditions, A1 for A and B, B1 for naming cosh x. The hyperbolic cosine is what this differential equation looks like when met without its name.

• 7 marks · find · stretch

  Find the general solution of y'' − 3y' + 2y = 4e^2x.

  Worked answer

  The auxiliary equation m² − 3m + 2 = 0 has roots 1 and 2, so the complementary function is Ae^x + Be^2x. Trying y = ke^2x for the particular integral fails, because that is already part of the complementary function and the left side collapses to zero. Multiply by x and try y = kxe^2x instead. Then y' = ke^2x(1 + 2x) and y'' = ke^2x(4 + 4x), so the left side is ke^2x[(4 + 4x) − 3(1 + 2x) + 2x] = ke^2x. Matching to 4e^2x gives k = 4, so the general solution is y = Ae^x + Be^2x + 4xe^2x. M1 for the auxiliary equation, A1 for the complementary function, B1 for spotting the duplication, M1 for trying kxe^2x, A1 for the substitution, A1 for k = 4, A1 for the general solution. Always check the trial function against the complementary function before differentiating. When the right-hand side duplicates a term of the complementary function, an extra factor of x is needed, and a repeated auxiliary root would require x².

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